If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=3…

If $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=5 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}$ are three vectors, then a vector which is perpendicular to $\vec{a}$ and $\vec{b} \times \vec{c}$ is
  1. $45 \hat{\mathrm{i}}-30 \hat{\mathrm{j}}+15 \hat{\mathrm{k}}$
  2. $3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
  3. $-30 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
  4. $-45 \hat{i}+30 \hat{j}+4 \hat{k}$

Solution

Vector perpendicular to $\vec{a}$ and $(\vec{b} \times \vec{c})$ will be $ \begin{aligned} & \vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} \\ & =(10)(3 \hat{j}+4 \hat{k})-(9)(5 \hat{i}+4 \hat{k}) \\ & =(-45 \hat{i}+30 \hat{j}+4 \hat{k}) \end{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

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